# Three red dice. Sum of any two of them must be 2,3 or 4. What is the probability of that?

I have a boardgame call Enemy Coast Ahead: The Dambusters in which enemy FLaK poses a major problem.

Under some circumstances the game "AI" throws three instead of two dice and if the sum of any two of them is within a certain interval a hit is scored. Furthermore if doubles are rolled and that gives a hit the doubles are re-rolled. If that is a hit but not a re-roll the hit is score and then it stops. If that is a hit and a double hit is score and another re-roll follows.

For starters I'd just like to know the probability of the sum of the two dice being 2,3 or 4 when rolling three dice.

As I see it(and I have not idea whether I right): Dice 1,2 and 3, denoted A, B and C can roll a lethal combbination as AB, BC or AC.

At Flak Level 1 the lethal range is 2-4. That gives 1/36 + 2/36 + 3/36=6/36 =1/6 for a Hit.

Using three dice and combinations AB, DC and AC just one of them is enough for a hit so since OR goes into it we get 1/6 for each 1/6 + 1/6 + 1/6 = 1/2.

Best Regards Peter

## 3 Answers

Let $$A$$ denote the event that $$X_1+X_2\in\{2,3,4\}$$.

Let $$B$$ denote the event that $$X_1+X_3\in\{2,3,4\}$$.

Let $$C$$ denote the event that $$X_2+X_3\in\{2,3,4\}$$.

With inclusion/exclusion and symmetry we find:$$P(A\cup B\cup C)=3P(A)-3P(A\cap B)+P(A\cap B\cap C)$$

Can you find these terms on RHS yourself?

If you calculate them correctly then your answer will be:$$3\cdot\frac6{36}-3\cdot\frac{14}{216}+\frac{11}{216}=\frac{77}{216}$$

• Argh! How do I read that P(A.... ? Is that "Set Theory"? – Peter Lageri Jan 3 at 12:42
• Is there something I should read to understand that? Its some 35 years since I went to school. – Peter Lageri Jan 3 at 12:50
• You should read $P(A)=P(X_1+X_2\text { equals }2, 3\text{ or }4)$. Here $A$ is indeed a set. A subset of the outcome space actually. It concerns the underlying theory for probability. If you want to endeepen then go looking for "probability space" on e.g. Wikipedia. – drhab Jan 3 at 12:52
• If this goes too deep then you better accept the answer of lulu. – drhab Jan 3 at 12:59
• Why 3P? Because there are three numbers that interest us? Was 6 in 6/36 found by brute force or calculated and if yes how? I would like to understand. – Peter Lageri Jan 3 at 13:08

You can't add the probability of non-mutually exclusive events like that. Suppose there were $$10$$ dice, so $$\binom {10}2=45$$ pairs. Your method would give an answer greater than $$1$$.

Best here is brute force counting.

Counting the winning triples requires attention because there is a lot of overlap and because not all of them have the same number of permutations.

$$6$$ permutations: $$\{1,2,3\}\quad \{1,2,4\}\quad\{1,2,5\}\quad\{1,2,6\}$$ $$\{1,3,4\}\quad \{1,3,5\}\quad\{1,3,6\}$$

$$3$$ permutations: $$\{1,1,2\}\quad \{1,1,3\}\quad\{1,1,4\}\quad\{1,1,5\}\quad \{1,1,6\}$$ $$\{1,2,2\}\quad \{2,2,3\}\quad \{2,2,4\}\quad\{2,2,5\}\quad\{2,2,6\}$$ $$\{1,3,3\}$$

$$1$$ permutation: $$\{1,1,1\}\quad \{2,2,2\}$$

Thus there are $$6\times 7+3\times 11+ 1\times 2=77$$ (ordered) winning triples. As there are $$6^3=216$$ unrestricted ordered triples, the answer is $$\boxed {\frac {77}{216}=.35648}$$

CAUTION: one problem with brute force enumeration is that it is quite error prone; it's hard to be sure you didn't miss a case or two. Check my list carefully. The first version of it left off two patterns and there might well be other errors that are still present.

• Ok, thank you. Why is it called 6 permutations and 3 permutations? – Peter Lageri Jan 3 at 12:17
• @PeterLageri I am writing down unordered triples and the $6,3,1$ refer to the number of ways to order them. $\{1,2,3\}$ for instance can be ordered $6$ ways, as $(1,2,3),(1,3,2),(2,1,3), (2,3,1), (3,1,2), (3,2,1)$. While $\{1,1,2\}$ can only be ordered in $3$ ways and $\{1,1,1\}$ only has $1$ order. – lulu Jan 3 at 12:21
• It is correct, lulu (unless. ...:-) – drhab Jan 3 at 12:28
• @drhab Seriously. I thought it would be a simple enumeration...can't think when I last made so many blunders. Glad to see the alternate computation you posted. – lulu Jan 3 at 12:34
• @Lulu. Thank you. – Peter Lageri Jan 3 at 12:37

Another way to brute force: $$\begin{array}{c|c|c|c} D_1&D_2&D_3&Total\\ \hline 1&1-3&1-6&18\\ 1&4-6&1-3&9\\ 2&1-2&1-6&12\\ 2&3-6&1-2&8\\ 3&1&1-6&6\\ 3&2&1-2&2\\ 3&3-6&1&4\\ 4-6&1&1-3&9\\ 4-6&2&1-2&6\\ 4-6&3&1&3\\ \hline &&&77 \end{array}$$ Hence the required probability is $$\frac{77}{6^3}$$.

• How do you calculate each linie? – Peter Lageri Jan 4 at 8:07
• Line 7 triples are: $(3,3,1);(3,4,1);(3,5,1);(3,6,1)$. – farruhota Jan 4 at 10:08
• Last line are: $(4,3,1);(5,3,1);(6,3,1)$. – farruhota Jan 4 at 10:09